1. Field of the Invention
The present invention relates generally to a power supply apparatus and a power connecting method, and more particularly to a power supply apparatus which, irrespective of a fault of a power source for a computer such as a fault-tolerant computer, continues to supply power to the computer and which is capable of exchanging the faulty power source as the hot line is kept to the energized state in terms of AC and DC.
2. Description of the Prior Art
FIG. 5 shows a circuit and arrangement of a basic portion of a conventional hot-line type power supply apparatus. In the illustration, numeral 1 represents a power unit making up the hot-line type power supply apparatus. In the internal circuit of the power unit 1, D1 to D4 respectively designate diodes, C1 and C2 respectively depict capacitors, L1 denotes an inductor, T1 indicates a pulse transformer, and TR1 is a switching transistor for converting into a pulse voltage a direct-current voltage developed across the capacitor C2. The diodes D2 and D3 make up a rectifier M, and the capacitor C1, the inductor L1 and others constitutes a smoothing circuit F.
Operation of the apparatus thus arranged will be described hereinbelow. In the power unit 1, an input rectified by a full-wave rectifier and ON-OFF controlled by the transistor TR1 is supplied to an input side of the transformer T1 to generate an output having a great amplitude, the output of the transformer T1 being rectified by the rectifier M and then smoothed by the smoothing circuit F so as to produce a direct current (DC) output whose voltage value is 5 [V], for example. In this case, an output line 30 for supplying the 5 [V] DC voltage is coupled in common to the output sides of the respective power units. Thus, even if one power unit malfunctions, the other power unit can compensate for this power unit trouble. Similarly, the input sides of the respective power units are coupled in common to an input line 31. Here, the diode D3 is a flywheel diode to obtain a rectified electric power. Although in FIG. 5 there are illustrated three power units, the 5 [V] DC output load can be covered by the total output capacity of two power units.
Now, assuming that one of the power units 1 malfunctions, difficulty is encountered to correctly generate the switching pulse at the secondary side 2 of the transformer T1 of the faulty power unit, whereby the voltage is lowered at the junction portion 3 between the inductor L1 and capacitor C1 making up the circuit F. Although the faulty power unit cannot output the 5 [V] DC because of the lowering of the voltage at the junction portion 3, the other two power units supply the 5 [V] DC therefor, thus allowing the continuous supply to the load.
On the other hand, the faulty power unit detects the lowering of the voltage at the junction portion 3 and gives information to the operator for exchange with a new one. The removal of the faulty power unit does not affect the 3 [V] output because current does not flow through an output terminal 4 line. Further, in inserting a new power unit in place of the faulty power unit, the voltage at the junction portion 3 is 0 [V] in the initial state and hence electric charge is not charged into the capacitor C1, while the reverse-current preventing diode D1 prevents the supply of current from the 5 [V] DC line to the capacitor C1 and hence the insertion of the new power unit does not affect the 5 [V] DC output. In response to the operation of this new power unit to restore the voltage at the junction portion 3, current directs through the diode D1 to the 5 [V] DC line.
In the conventional hot-line type power supply apparatus thus arranged, since the reverse-current preventing diode D1 is always connected in series to the output portion of the power supply, when the forward-direction voltage drop is taken to be V.sub.F, the power loss is taken as P and the output current of the power unit is I.sub.F, they take the following relation and there is a problem that the conversion efficiency of the power supply lowers and generates heat. EQU P=V.sub.F .times.I.sub.F
For example, when I.sub.F =100 A and V.sub.F =1 V, the power loss P becomes 100 W, thus resulting in a great problem.